When solving the heat equation on say $\mathbb{R}$ (or $[0,2\pi]$, whichever is easier to talk about) we are posing Cauchy data on the surface $t=0$. My understanding is that $t=$constant are precisely the characteristic surfaces of the heat equation.

I realize this question may seem elementary to those who know the answer, but I'm confused as to how this makes sense. I understand how to solve the heat equation using Fourier series (on $[0,2\pi]$) or the fundamental solution on $\mathbb{R}$ but I thought we were not able to pose Cauchy data on characteristic surfaces? Is the point here that we are not solving the equation using the characteristics? Shouldn't $u_t$ be automatically specified in terms of what $u_{xx}(t=0,x)$ is?

3 Answers
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The concept of a non-characteristic surface for a PDE or a system of PDE's is useful primarily for only establishing the existence and uniqueness of real analytic or formal power series solutions to the initial value problem using the Cauchy-Kovalevsky theorem.

The generalization of this to the smooth category is the class of hyperbolic PDE's, where you need the initial hypersurface to be more than characteristic. It has to be space-like.

Parabolic equations are a set of PDE's for which the initial value problem in time is well-posed only in the smooth category and not in the real analytic category. Indeed, if you try to apply the Cauchy-Kovalevsky theorem, the $t = c$ hypersurface is characteristic. From the point of view of this theorem, the initial value problem for the standard heat operator is well-posed only for hypersurfaces that are noncharacteristic with respect to the space-like Laplacian. The time derivative is lower order and does not even appear in the symbol.

However, in any useful application of parabolic equations, you want smooth solutions to the initial value problem in time. The explicit formula for the heat kernel shows that the solution is not necessarily real analytic in time. For parabolic equations, the study of solutions that are real analytic or have a power series expansion in the time variable is of little interest. You want to use a weaker category of solutions.

Oh Deane, how is it that you always manage to explain things so much better than I can? (Removed my more or less duplicate, but inferior answer.)
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Willie WongSep 11 '10 at 12:42

I just wanted to ask Thanks! I guess I'm surprised that we even can do it though. When focusing in on first order equations there are compatibility issues. For instance I can't pose Cauchy data for the equation $\partial_x u =0$ on $\mathbb{R}$ unless my Cauchy data is constant. For the wave equation I need to make sure that my Cauchy data does not conflict with my boundary data (for instance specifying both $u_t$ and $u_x$ on the boundary $x=0$ for the semi-infinite line problem would be a problem).
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DorianSep 11 '10 at 14:30

Is there some sort of equivalent trouble for the heat equation? It seems like for parabolic equations I can pose ANY cauchy data I want as long as it is smooth enough and I get a solution. I just find this surprising considiner all of the other examples one encounters, there are noticable problems with consistency of cauchy data versus boundary data. Are parabolic equations free of this?
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DorianSep 11 '10 at 14:31

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Willie, according to my calculations I have had a 26 year head start on you. If one adjusts for that you appear to be well ahead of me.
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Deane YangSep 11 '10 at 14:36

Dorjan, your questions are too vague for me to understand and answer. Certainly, if you are trying to solve an initial value problem on a space domain with boundary, you have to worry about the boundary data, just as you do for a hyperbolic PDE. For the classical heat and wave equations, this is explained pretty well in most PDE or mathematical physics texts. As for the equation $\partial_xu = 0$, it is definitely not parabolic, so there is no reason to expect the characteristic initial value problem to be well-posed. The behavior of a PDE depends in general on more than the top order terms.
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Deane YangSep 11 '10 at 16:28

Your question originates in a confusion about the words Cauchy problem.

When you have a linear PDE of order $k$, and a hypersurface $S$, it seems natural to try to solve the Cauchy problem with data posed on $S$. Cauchy problem means here that you give the unknown $u$ and the $k-1$ first normal derivatives (whence $k$ scalar data). When you do that, you immediately have all the derivatives of $u$ over $S$ up to order $k-1$. Then you ask yourself whether the PDE gives you the rest of the jet, by induction on the order of derivatives. When the answer to this question is positive, you say that $S$ is non-characteristic.

When the answer is negative, you realize that the PDE actually imposes non-trivial linear relations between $u,Du,\ldots,D^{k-1}u$ over $S$. In other words, you are not free to choose the Cauchy data over $S$. This means that you have to start with less many data (less than $k$).

This is what happens for the Heat equation with $S$ the set $t=0$. The Cauchy problem, in the original sense of these words, would be to give both $u$ and $u_t$ at initial time, because $k=2$. But $S$ is characteristic and therefore one may not give two scalar data. In the case of the Heat equation, it is not hard to see that what is important is the partial order of the equation with respect to the direction normal to $S$ (the time). This order is one, whence only one initial data. This is a general principle.

You can pose Cauchy data anywhere you want. The question is whether solutions exist and are unique. If you just look at first order linear equations, you immediately see that at least one of existence and uniqueness will fail. For more general systems of PDE, you have the theorem of Cauchy-Kovalevskaya which guarantees existence and uniqueness of solution for analytic initial data posed on noncharacteristic surfaces, conditional on the fact that the top order transversal derivative in the principal symbol is nondegenerate.
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Willie WongSep 11 '10 at 11:06

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Sometimes it is possible to pose data on characteristic surfaces, but to establish existence and uniqueness of solutions for them requires additional hard work not covered by standard nonsense. See, e.g. the papers of Cognac and Dossa on the initial value problem for the wave equation with data prescribed on a characteristic cone.
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Willie WongSep 11 '10 at 11:08